Question

in the arithmetic series 2,5,8... up to 50 terms and 3,5,7,9.... up to 60 terms find how many terms are identical

Answer 1

AnswEr:

Let the mth term of the first A.P. be equal to the nth term of the second A.P. Then,

$\qquad \tt \: 2 + (m - 1) \times 3 = 3 + (n - 1) \times 2 \\ \\ \tt \rightarrow \: 3m - 1 = 2n + 1 \\ \\ \tt \rightarrow \:3m = 2n + 2 \\ \\ \tt \rightarrow \: \frac{m}{2} = \frac{n + 1}{3} = k \: \: \: \: (say) \\ \\ \rightarrow \tt \: m = 2k \: \: \: and \: \: \: n = 3k - 1 \\ \\ \tt \Rightarrow \:2k \leqslant 50 \: \: \: and \: \: \: 3k - 1 \leqslant 60 \\ \\ \tt \Rightarrow \:k \leqslant 25 \: \: \: and \: \: \: k \leqslant 20 \frac{1}{3} \\ \\ \tt \Rightarrow \:k \leqslant 20 \\ \\ \tt \Rightarrow \:k = 123...20$

Corresponding to each value of k, we get a pair of identical terms.

• Hence, there are 20 identical terms in the two A.P.'s.